Improved Outerplanarity Bounds for Planar Graphs

(1) We chose root-vertex $r$ so that it is not a cutvertex; therefore $G\setminus L_{0}=G\setminus r$ is connected and there is only one node that is a child of $R$.

(2) For edge $(y,z)$, let $i$ be the smallest index for which layer $L_{i}$ contains $y$ or $z$. If both $y,z$ are in $L_{i}$, then $y,z$ belong to the same node since they are in one connected component. If one of them (say $y$) is not in $L_{i}$, then $y\in L_{i+1}$, so the edge is directed $y\rightarrow z$ and $Y$ becomes a child of $Z$ since the edge $(y,z)$ ensures that $y$ is in the connected component that defined $Z$.

(3) Recall that node $N$ corresponds to a connected component $K_{N}$ of the graph obtained by deleting some of the levels. Since $N$ is not a leaf, subgraph $K_{N}$ has at least one vertex $v$ not on the outerface. Therefore, the outerface of $K_{N}$ contains a cycle that has $v$ inside and $r$ outside (since $N\neq R$). This cycle is a fence and all its vertices belong to $V(N)$. ∎

Observe first that since we start with the same root-vertex and outerface, and only add directed edges, color=blue!50!white]FYI: We used to assume that $H$ is minimal, but I dont want to waste effort on discussing how to find this efficiently. So downgraded this to ‘add only directed edges’, which is easy to achieve. the layers $L_{0},\dots,L_{k}$ are exactly the same in both $G$ and $H$. Assume $H$ is obtained by adding just one edge $e=(y,z)$ (the full proof is then by induction on the number of added edges). Observe that $y,z$ belong to one face $F$ of $G$ since we can add edge $(y,z)$ to $G$ while staying planar. Since $G$ is connected, so is the boundary of $F$. For any face the vertices belong to at most two consecutive layers by definition of peels; for the specific face $F$ the vertices belong to exactly two consecutive layers (say $L_{i}$ and $L_{i+1}$) since they include $y,z$ which are not on the same layer by construction of $H$. So there exists in $G$ a path $\pi$ (along the boundary of $F$) that connects $y$ and $z$ and has all vertices in $L_{i}$ and $L_{i+1}$.

With this, the connected components that define the tree of peels are exactly the same for both graphs. To see this, observe that when we have removed $L_{0}\cup\dots\cup L_{h-1}$ for some $h\leq i$, then edge $(y,z)$ provides no connectivity among components that we did not have via path $\pi$ instead. Once we have removed $L_{i}$, one of $y,z$ (and hence the added edge) has been removed from the graph, and so again does not add any connectivity. ∎

Recall that there are directed edges $s_{j-1}\rightarrow s_{j}$ and $z_{j-1}\rightarrow z_{j}$ by $j>0$ and that $s_{j-1},z_{j-1}\in A_{j-1}$. Write $G_{j}$ for the graph induced by $V(A_{j})$, and observe that $V(A_{j-1})$ must all belong to one inner face $F$ of $G_{j}$ since it bounds a connected component of the subgraph of $G$ where the peels up to $A_{j}$ have been removed. Furthermore, $s_{j},z_{j}$ have neighbours both strictly inside and strictly outside $F$ (due to their outgoing edges). So there must be a simple cycle $C$ along the boundary of $F$ that contains both $s_{j}$ and $z_{j}$. Walking along the shorter side of $C$ hence gives a walk from $s_{j}$ to $z_{j}$ of length at most $\lfloor|C|/2\rfloor\leq\lfloor|V(A_{j})|/2\rfloor$. By $j<i^{*}$ the stopping-condition did not hold at $A_{j}$, so this implies $\lfloor|V(A_{j})/\rfloor>\xi_{i}$ and the result holds by integrality. ∎

We are done if $A_{0}$ is the root $R$ or its child or a grandchild of $R$, for then we can connect $s_{0}$ and $t_{0}$ with a path of length at most 4 by following directed edges until we reach root-vertex $r$. So assume that $A_{0}$ has a parent $P$, grand-parent $G_{1}$ and great-grandparent $G_{2}$, therefore ${\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}\mathbf{a}}(A_{0})\geq|V(P)|+|V(G_{1})|+|V(G_{2})|\geq 3+3+1=7$ since $P$ and $G_{1}$ are internal nodes. Hence $\lceil\sqrt{{\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}\mathbf{a}}(A_{0})}\rceil\geq 3$ and the desired upper bound is $2\lceil\sqrt{{\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}\mathbf{a}}(A_{0})}\rceil-2\geq 4$. For ease of writing define $\beta=\lceil\sqrt{{\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}\mathbf{a}}(A_{0})}\rceil-3\geq 0$, so the desired upper bound becomes $2\beta+4$. Apply the detour-method at $A_{0}$, using $\xi_{i}=2\beta+4-2i$ for $i\geq 0$. If the method returns successfully at index $i^{*}$ with paths $\tau_{s},\tau_{z},\sigma$, then combining the three paths gives $d(s_{0},t_{0})\leq|\tau_{s}|+|\tau_{z}|+|\sigma|\leq 2i^{*}+2\beta+4-2i^{*}$ as desired.

So now assume for contradiction that the detour-method fails, and let $A_{i},s_{i},z_{i}$ (for $i=0,\dots,\ell$) be the nodes and vertices that it used; we have $A_{\ell}=R$ since the method can only fail at the root. Since the detour-method did not succeed at $R$, color=blue!50!white]FYI: This needed a major rewrite; Obs.3.4 doesn’t hold for the root note. we did not have a path of length at most $\xi_{\ell}$ from $s_{\ell}$ to $z_{\ell}$. But $|R|=1$, so $s_{\ell}=r=z_{\ell}$ are connected by a path of length 0. So $0>\xi_{\ell}=2\beta+4-2\ell$ or $\ell>\beta+2$. Now bound the sizes of $V(A_{1}),\dots,V(A_{\ell})$ as follows:

• •

  For $i=1,\dots,\beta+2$, Observation 3.4 implies $|V(A_{i})|\geq 2(\xi_{i}+1)=4\beta+10-4i>3\beta+8-4i$.

• •

  For $i=\beta{+}1$, this bound becomes $|V(A_{i})|\geq 4\beta+10-4(\beta{+}1)=6$.

• •

  For $i=\beta{+}2$, we also know $|V(A_{i})|\geq 3$ since $A_{i}$ is not the root by $\ell>\beta+2$.

• •

  Root $A_{\ell}$ stores one vertex $r$.

We therefore have a contradiction:

∎

Let $\Pi$ be the path from $S$ to $Z$ in ${\cal T}$. Every interior node $N$ of $\Pi$ belongs to the same subtree ${\cal T}^{\prime}$ of ${\cal T}\setminus S$, and satisfies $|V(N)|\geq g$ since it is neither leaf nor root. Node $Z\neq S$ also belongs to ${\cal T}^{\prime}$ and $|V(Z)|\geq 1$. Altogether therefore $\tfrac{n}{2}\geq|V({\cal T}^{\prime})|\geq g(|\Pi|{-}1)+1$ or $|\Pi|\leq\tfrac{n-2}{2g}+1$, which implies $d_{{\cal T}}(S,Z)=|\Pi|\leq\delta$ by integrality. ∎

Figures 2 and 3(a) illustrate this proof. Let $Z$ be the node that stores $z$, and let $A_{0}$ be the least common ancestor of $S$ and $Z$; this is a strict ancestor of $S$ since $Z\in{\cal T}_{R}$. Follow directed edges from $s$ to some vertex $s_{0}\in V(A_{0})$; the resulting path $\pi_{s}$ has length $d_{\cal T}(S,A_{0})$ since every directed edge gets us closer to the root. Likewise we can get a path $\pi_{z}$ of length $d_{\cal T}(Z,A_{0})$ from $z$ to some node $z_{0}\in V(A_{0})$ (possibly $z_{0}=z$).

Now apply the detour-method with $A_{0},s_{0},z_{0}$, using $\xi_{i}=g{-}1$. This will always exit with ‘success’ at a node $A_{i^{*}}$ that is not the root, because any two vertices stored at the child of the root have outgoing edges towards the root-vertex $r$ and hence distance $2\leq g{-}1$. Let $\tau_{s},\tau_{z},\sigma$ be the paths, then $\pi:=s{\overset{\pi_{s}}{\>\text{{\char 58\relax\char 59\relax} }\!}}s_{0}{\overset{\tau_{s}}{\>\text{{\char 58\relax\char 59\relax} }\!}}s_{i^{*}}{\overset{\sigma}{\>\text{{\char 58\relax\char 58\relax} }\!}}z_{i^{*}}{\overset{\tau_{z}}{\raisebox{4.30554pt}{\rotatebox{180.0}{$\,\>\text{{\char 58\relax\char 59\relax} }\!\,$}}}}z_{0}{\overset{\pi_{z}}{\raisebox{4.30554pt}{\rotatebox{180.0}{$\,\>\text{{\char 58\relax\char 59\relax} }\!\,$}}}}z$ has length $d_{{\cal T}}(S,Z)+2i^{*}+|\sigma|$.

To bound $|\pi|$, consider the path $\Pi$ from $S$ to $Z$ in ${\cal T}$, which goes through $A_{0}$, and let $A_{1},\dots,A_{i^{*}}$ be the ancestors of $A_{0}$ that were visited by the detour-method. If $i^{*}>0$, then by Obs. 3.4 we have $|V(A_{j})|\geq 2(\xi_{j}{+}1)=2g$ for $j=1,\dots,i^{*}{-}1$, node $A_{i^{*}}$ and the interior nodes of $\Pi$ store at least $g$ vertices each which nodes $Z$ and $R$ store at least one vertex. Therefore

hence $|\Pi|+2i^{*}\leq\tfrac{n{-}4}{2g}+2$ which is at most $\delta{+}1$ by integrality. If $i^{*}=0$ then $|\Pi|+2i^{*}\leq\delta$ by Obs. 4.1. Either way $|\pi|=|\Pi|+2i^{*}+|\sigma|\leq\delta+1+(g{-}1)$. ∎

See also Figure 3(b). Use directed edges to find a path $\pi_{z}$ of length $d_{\cal T}(Z,S)\leq\delta$ from $z$ to some vertex $z_{0}\in V(S)$. We know $d_{H}(s,z_{0})\leq\max\{2\lfloor\sqrt{{\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}\mathbf{a}}(S)}\rfloor{-}2,4\}\leq 2g-2$ by Lemma 1. Combining the paths gives the desired length: $d_{H}(z,s)\leq d_{H}(z,z_{0})+d_{H}(z_{0},s)\leq\delta+2g-2$. ∎

See Figure 3(c). Use directed edges to go from $z$ to a vertex $z_{D}\in V(D)$ along a path $\pi_{z}$ of length $d_{\cal T}(Z,D)$. Find a path $\sigma$ of length at most $\lfloor\tfrac{|V(D)|}{2}\rfloor$ to connect $z_{D}$ to $s_{D}$, and then follow $d_{\cal T}(D,S)$ directed edges to get to $s$. Combining the paths gives the desired length since $d_{\cal T}(Z,D)+d_{\cal T}(D,S)=d_{\cal T}(Z,S)$.∎

Recall that $\delta=\lfloor\tfrac{n-2}{2g}\rfloor+1$ is an integer and observe that $\delta\geq\tfrac{n-1}{2g}$. We also know that

Therefore

which yields the result by integrality. ∎

The method is exactly the same as in the proof of Claim 4.3 (walk from $z$ to a vertex $z_{0}\in V(S)$ and apply Lemma 1 to connect $z_{0}$ to $s$, see also Figure 3(b)), but the analysis is different. Since $Z$ is not deep, we have $d_{H}(z,z_{0})=d_{\cal T}(Z,S)\leq\theta{-}1\leq\delta{-}2\lceil\sqrt{{\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}\mathbf{a}}(S)}\rceil{+}2g$. By ${\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}\mathbf{a}}(S)>g^{2}\geq 9$ we have $d_{H}(z_{0},s)\leq 2\lceil\sqrt{{\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}\mathbf{a}}(S)}\rceil{-}2$ and so $d_{H}(z,s)\leq d_{H}(z,z_{0})+d(z_{0},s)\leq\delta+2g-2$. ∎

Let $X$ be the least common ancestor of all deep nodes. This is a descendant of $S$ as well (possibly $X=S$), so enumerate the path from $S$ to $X$ as $S{=}D_{0},D_{1},\dots,\allowbreak D_{k}{=}X$ for some $k\geq 0$. We are done if $|V(D_{i})|\leq 2g-1$ for some $0\leq i\leq k$, so assume (for contradiction) that the $k$ nodes $D_{1},\dots,D_{k}$ store at least $2g$ vertices each. If $X$ were deep (so $k\geq\theta$) then $D_{1},\dots,D_{k}$ would store at least $\theta\cdot 2g\geq n-{\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}\mathbf{a}}(S)-2g$ vertices. We also store ${\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}\mathbf{a}}(S)$ vertices at strict ancestors at $S$, $|V(S)|$ vertices at $S$ and at least $4g-|V(S)|+1$ deep vertices, in total hence more than $n$, impossible.

So $X$ is not deep, see also Figure 3(d). Since $X$ is the least common ancestor of deep descendants, therefore it must have at least two children $X^{1},X^{2}$ that are ancestors of deep descendants. For $j=1,2$, enumerate the path from $X^{j}$ to a deep descendant of $X^{j}$ as $D^{j}_{k+1},\dots,D^{j}_{\theta}$. Then for $i=k{+}1,\dots,\theta{-}1$ node $D_{i}^{j}$ is not a leaf and stores at least $g$ vertices, so $|V(D_{i}^{1})|+|V(D_{i}^{2})|\geq 2g$. So for $i=1,\dots,\theta-1$ we can find $2g$ vertices stored at not-deep descendants of distance $i$ from $S$. In total these not-deep strict descendants of $S$ hence store at least $(\theta{-}1)2g\geq n-{\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}\mathbf{a}}(S)-4g$ vertices. As above therefore more than $n$ vertices are stored in the tree of peels, impossible. ∎

Compute layers $L_{0},L_{1},\dots$, augmentation $H$, and tree of peels ${\cal T}$ as in Section 3. Find separator-node $S$ and compute $|V(S)|$, ${\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}\mathbf{a}}(S),\theta$, and the number of deep vertices. If $|V(S)|\geq 4g{-}2$ and there are at least $4g{-}|V(S)|{+}1$ deep vertices, then arbitrarily fix a deep node $Z$ and find node $D$ of Claim 4.7 by walking from $S$ towards $Z$ until we encounter a node that stores at most $2g-1$ vertices. In all other cases set $D:=S$. Pick $s$ as in Claim 4.4 applied to $D$. Each of these steps takes linear time (we elaborate on this in Section 4.1).

Applying various cases we show that $d_{H}(s,z)\leq\delta+2g-2$ for all $z$ (which implies the result by $\delta=\lfloor\tfrac{n-2}{2g}\rfloor+1$). This holds for all $z\in V({\cal T}_{R})$ by Claim 4.2, so consider a vertex $z$ stored at a descendant $Z$ of $S$. The bound holds by Claim 4.3 if ${\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}\mathbf{a}}(S)\leq g^{2}$ and by Corollary 1 if $|V(S)|\leq 4g-3$, so assume neither. If $Z$ is not deep, then apply Claim 4.6. If $Z$ is deep and there are at least $4g-|V(S)|+1$ deep vertices, then combine Claim 4.7 with Claim 4.4 to get the bound. The only remaining case is that ${\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}\mathbf{a}}(S)>g^{2}$, $|V(S)|\geq 4g-2$, $Z$ is deep, and at most $4g-|V(S)|$ vertices are deep (so $|V(S)|\leq 4g-1$ since there is a deep vertex at $Z$). In this case, $d_{\cal T}(Z,S)=\theta$, for otherwise $Z$’s parent would also be deep and store $g\geq 3\geq 4g-|V(S)|+1$ deep vertices. Also $\theta\leq\delta-2\lceil\sqrt{{\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}\mathbf{a}}(S)}\rceil+2g+1\leq\delta-1$ by $\lceil\sqrt{{\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}\mathbf{a}}(S)}\rceil\geq g+1$. We used $D=S$ and picked $s\in V(S)$ as in Claim 4.4, so this gives $d_{H}(z,s)\leq d_{T}(Z,S)+\lceil\tfrac{V(S)}{2}\rceil\leq\delta-1+2g-1$ as desired. ∎